\( \int\displaystyle\frac{dx}{(a^2-x^2)^{3/2}} \) is equal to: (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{x}{a^2\sqrt{a^2-x^2}}+C \)
The expression \( a^2-x^2 \) suggests the substitution \( x=a\sin\theta \).
\( \displaystyle x=a\sin\theta,\quad dx=a\cos\theta\,d\theta \)
\( \displaystyle (a^2-x^2)^{3/2}=(a^2\cos^2\theta)^{3/2} \)
\( \displaystyle =a^3\cos^3\theta \)
\( \displaystyle \int\frac{dx}{(a^2-x^2)^{3/2}}=\int\frac{a\cos\theta}{a^3\cos^3\theta}\,d\theta \)
\( \displaystyle =\frac{1}{a^2}\int\sec^2\theta\,d\theta \)
\( \displaystyle =\frac{1}{a^2}\tan\theta+C \)
Return to x. Since \( \sin\theta=\frac{x}{a} \), \( \cos\theta=\frac{\sqrt{a^2-x^2}}{a} \), so
\( \displaystyle \tan\theta=\frac{x}{\sqrt{a^2-x^2}} \)
\( \displaystyle \int\frac{dx}{(a^2-x^2)^{3/2}}=\frac{x}{a^2\sqrt{a^2-x^2}}+C \)
Shortcut: Differentiate Option (2) instead of integrating. Write it as \( \frac{1}{a^2}\,x(a^2-x^2)^{-1/2} \):
\( \displaystyle \frac{d}{dx}\left[x(a^2-x^2)^{-1/2}\right]=(a^2-x^2)^{-1/2}+x^2(a^2-x^2)^{-3/2} \)
\( \displaystyle =\frac{(a^2-x^2)+x^2}{(a^2-x^2)^{3/2}} \)
\( \displaystyle =\frac{a^2}{(a^2-x^2)^{3/2}} \)
Dividing by \( a^2 \) gives the integrand.
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{a^2}\sqrt{a^2-x^2}+C \) — Incorrect. Its derivative is \( -\frac{x}{a^2\sqrt{a^2-x^2}} \), not the integrand.
Option (3) \( \displaystyle\frac{x}{a^2\sqrt{a^2+x^2}}+C \) — Incorrect. This is the answer for \( \int\frac{dx}{(a^2+x^2)^{3/2}} \) (substitution \( x=a\tan\theta \)); the sign inside the root is wrong here.
Option (4) \( \displaystyle\frac{2}{5a^2}(a^2-x^2)^{5/2}+C \) — Incorrect. The power rule cannot be used on a power of \( a^2-x^2 \) without the factor \( -2x \); differentiating this option gives \( -\frac{2x}{a^2}(a^2-x^2)^{3/2} \).